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          数组元素与下标
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        <p>&ensp;&ensp;&ensp;&ensp;在有关数组的算法题中，经常会对数组的元素限定范围，通常为<code>1 ~ nums.size()</code>，这样的题目往往可以把数组元素和其下标联系起来。</p>
<span id="more"></span>

<h3 id="一、寻找重复数"><a href="#一、寻找重复数" class="headerlink" title="一、寻找重复数"></a><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/find-the-duplicate-number/">一、寻找重复数</a></h3><h4 id="1-1题目"><a href="#1-1题目" class="headerlink" title="1.1题目"></a>1.1题目</h4><p>&ensp;&ensp;&ensp;&ensp;给定一个包含 <code>n + 1</code> 个整数的数组 <code>nums</code> ，其数字都在 <code>1</code> 到 <code>n</code> 之间（包括 <code>1</code> 和 <code>n</code>），可知至少存在一个重复的整数。</p>
<p>&ensp;&ensp;&ensp;&ensp;假设 <code>nums</code> 只有 <strong>一个重复的整数</strong> ，找出 <strong>这个重复的数</strong> 。</p>
<p><strong>示例 1：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：nums &#x3D; [1,3,4,2,2]</span><br><span class="line">输出：2</span><br></pre></td></tr></table></figure>

<p><strong>示例 2：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：nums &#x3D; [3,1,3,4,2]</span><br><span class="line">输出：3</span><br></pre></td></tr></table></figure>

<p><strong>示例 3：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：nums &#x3D; [1,1]</span><br><span class="line">输出：1</span><br></pre></td></tr></table></figure>

<p><strong>示例 4：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：nums &#x3D; [1,1,2]</span><br><span class="line">输出：1</span><br></pre></td></tr></table></figure>

<p><strong>提示：</strong></p>
<ul>
<li>  <code>2 &lt;= n &lt;= 3 * 104</code></li>
<li>  <code>nums.length == n + 1</code></li>
<li>  <code>1 &lt;= nums[i] &lt;= n</code></li>
<li>  <code>nums</code> 中 <strong>只有一个整数</strong> 出现 <strong>两次或多次</strong> ，其余整数均只出现 <strong>一次</strong></li>
</ul>
<h4 id="1-2思路与代码"><a href="#1-2思路与代码" class="headerlink" title="1.2思路与代码"></a>1.2思路与代码</h4><p>&ensp;&ensp;&ensp;&ensp;先说一下自己一开始的思路，本来想先排序，再遍历，但这样明显低估了本题，时间复杂度太大。后来联想到了<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/first-missing-positive/">leetcode: 41</a>，由于面试时遇到过此题，对其印象深刻，故写了相似的算法：</p>
<ol>
<li> 把数组中的元素<strong>归位</strong>，比如数组中的 1 放在数组第1个位置， 2 放在数组第2个位置……</li>
<li> 当 <code>nums[i] != i + 1</code>时，若<code>nums[i] == nums[nums[i] - 1]</code>，说明<code>nums[i]</code>就是要找的数</li>
<li> 当 <code>nums[i] != i + 1</code>时，若<code>nums[i] != nums[nums[i] - 1]</code>，交换<code>nums[i]</code>和<code>nums[nums[i] - 1]</code></li>
<li> 必须确保再找到目标之前，<code>nums[i] = i + 1</code>（所以交换的过程可能是多个）</li>
</ol>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span></span><br><span class="line"><span class="class">&#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">findDuplicate</span><span class="params">(vector&lt;<span class="keyword">int</span>&gt;&amp; nums)</span> </span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; nums.<span class="built_in">size</span>(); ++i)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">if</span> (nums[i] == i + <span class="number">1</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">continue</span>;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">while</span> (nums[i] != i + <span class="number">1</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">if</span> (nums[i] == nums[nums[i] - <span class="number">1</span>])</span><br><span class="line">                &#123;</span><br><span class="line">                    <span class="keyword">return</span> nums[i];</span><br><span class="line">                &#125;</span><br><span class="line">                <span class="keyword">else</span></span><br><span class="line">                &#123;</span><br><span class="line">                    <span class="built_in">swap</span>(nums[i], nums[nums[i] - <span class="number">1</span>]);</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> nums[nums.<span class="built_in">size</span>() - <span class="number">1</span>];</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<p>&ensp;&ensp;&ensp;&ensp;运行之后，发现时间复杂度还是太高，仅仅比排序好了一点。看了一眼评论区，在大佬的启发下写了快慢指针的解题方法。</p>
<p>&ensp;&ensp;&ensp;&ensp;快慢指针思想：<code>fast</code> 和 <code>slow</code> 是指针，<code>nums[slow]</code> 表示取指针对应的元素注意 <code>nums</code> 数组中的数字都是在 <code>1</code> 到 <code>n</code> 之间的(在数组中进行游走不会越界)，因为有重复数字的出现，所以这个游走必然是成环的，环的入口就是重复的元素，即按照寻找链表环入口的思路来做。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span></span><br><span class="line"><span class="class">&#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">findDuplicate</span><span class="params">(vector&lt;<span class="keyword">int</span>&gt;&amp; nums)</span> </span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> slow = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">int</span> fast = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">while</span> (<span class="number">1</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            slow = nums[slow];</span><br><span class="line">            fast = nums[nums[fast]];</span><br><span class="line">            <span class="keyword">if</span> (slow == fast)</span><br><span class="line">            &#123;</span><br><span class="line">                fast = <span class="number">0</span>;</span><br><span class="line">                <span class="keyword">while</span> (nums[slow] != nums[fast])</span><br><span class="line">                &#123;</span><br><span class="line">                    slow = nums[slow];</span><br><span class="line">                    fast = nums[fast];</span><br><span class="line">                &#125;</span><br><span class="line">                <span class="keyword">return</span> nums[slow];</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h3 id="二、找到所有数组中消失的数字"><a href="#二、找到所有数组中消失的数字" class="headerlink" title="二、找到所有数组中消失的数字"></a><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/find-all-numbers-disappeared-in-an-array/">二、找到所有数组中消失的数字</a></h3><h4 id="2-1题目"><a href="#2-1题目" class="headerlink" title="2.1题目"></a>2.1题目</h4><p>&ensp;&ensp;&ensp;&ensp;给定一个范围在 1 ≤ a[i] ≤ <em>n</em> ( <em>n</em> = 数组大小 ) 的 整型数组，数组中的元素一些出现了两次，另一些只出现一次。</p>
<p>&ensp;&ensp;&ensp;&ensp;找到所有在 [1, <em>n</em>] 范围之间没有出现在数组中的数字。</p>
<p>&ensp;&ensp;&ensp;&ensp;能在不使用额外空间且时间复杂度为*O(n)*的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内。</p>
<p><strong>示例:</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">输入:</span><br><span class="line">[4,3,2,7,8,2,3,1]</span><br><span class="line">输出:</span><br><span class="line">[5,6]</span><br></pre></td></tr></table></figure>

<h4 id="2-2思路与代码"><a href="#2-2思路与代码" class="headerlink" title="2.2思路与代码"></a>2.2思路与代码</h4><p>&ensp;&ensp;&ensp;&ensp;数组的元素有范围，便可以将它们与自己的下标联系起来。这里采用力扣评论区的一个方法：将所有正数作为数组下标，置对应数组值为负值。那么，仍为正数的位置即为（未出现过）消失的数字。</p>
<figure class="highlight tex"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">举个例子：</span><br><span class="line">原始数组：[4,3,2,7,8,2,3,1]</span><br><span class="line">重置后为：[-4,-3,-2,-7,8,2,-3,-1]</span><br></pre></td></tr></table></figure>

<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span></span><br><span class="line"><span class="class">&#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="keyword">int</span>&gt; <span class="title">findDisappearedNumbers</span><span class="params">(vector&lt;<span class="keyword">int</span>&gt;&amp; nums)</span> </span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> n = nums.<span class="built_in">size</span>();</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; n; ++i)</span><br><span class="line">        &#123;</span><br><span class="line">            nums[<span class="built_in">abs</span>(nums[i]) - <span class="number">1</span>] = -<span class="built_in">abs</span>(nums[<span class="built_in">abs</span>(nums[i])<span class="number">-1</span>]);</span><br><span class="line">        &#125;</span><br><span class="line">        vector&lt;<span class="keyword">int</span>&gt; ans;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; n; ++i)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">if</span> (nums[i] &gt; <span class="number">0</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                ans.<span class="built_in">push_back</span>(i + <span class="number">1</span>);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> ans;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>


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